Minimum Speed to Arrive on Time

10 Prerequisite LeetCode Problems

For “1870. Minimum Speed to Arrive on Time”, the following are a good preparation:

“704. Binary Search” - This problem helps in understanding the basic concept of binary search which is an integral part of solving the main problem.
“278. First Bad Version” - Similar to the main problem, this requires a binary search approach to find the earliest possible bad version, similar to finding the minimum speed.
“33. Search in Rotated Sorted Array” - Understanding how binary search works on modified sorted arrays is a good preparation for the main problem.
“153. Find Minimum in Rotated Sorted Array” - This problem provides good practice for implementing binary search on a sorted and rotated array, similar to searching for a particular speed in the main problem.
“35. Search Insert Position” - This problem involves finding the position where an element should be, similar to finding the speed.
“34. Find First and Last Position of Element in Sorted Array” - This problem helps in understanding how to find the first occurrence of an element using binary search.
“162. Find Peak Element” - This problem requires an understanding of how to find a peak element in an array which is crucial to understanding how binary search works in a variety of scenarios.
“167. Two Sum II - Input array is sorted” - The problem deals with a sorted array and finding a pair of numbers, similar to finding a specific speed.
“852. Peak Index in a Mountain Array” - This problem helps in understanding how to find the maximum element in an array, similar to finding the minimum speed.
“349. Intersection of Two Arrays” - The solution involves sorting and binary searching, the same techniques used to solve the main problem.

These involve binary search and sorting techniques which are integral to solving the main problem.

Problem Classification

This can be classified as “Combinatorial Optimization”, as we’re tasked with finding an optimal (minimum) speed to ensure reaching the office on time.

‘What’ Components:

We have a floating-point number ‘hour’, indicating the maximum time available to reach the office.
We have an array ‘dist’ of integers with ’n’ elements, each representing the distance for the ‘i-th’ train ride in the journey.
Each train ride can only begin at an integer hour mark. If a train ride ends at a non-integer hour, waiting until the next integer hour is required before starting the next ride.
We need to find the minimum positive integer speed for all the trains to ensure reaching the office within the given time ‘hour’. If it’s impossible to reach on time, return -1.

We can classify this as a “Search Problem”, specifically a “Binary Search” problem, as we are trying to find the minimum speed in a specific range (1 to 107) which satisfies the condition of reaching the office on time. The waiting times and sequential ordering of the train rides make it a more complex variant, which requires effective problem-solving skills.

Clarification Questions

What are the clarification questions we can ask about this problem?

Problem Analysis and Key Insights

What are the key insights from analyzing the problem statement?

Problem Boundary

What is the scope of this problem?

How to establish the boundary of this problem?

Distilling the Problem to Its Core Elements

Can you identify the fundamental concept or principle this problem is based upon? Please explain. What is the simplest way you would describe this problem to someone unfamiliar with the subject? What is the core problem we are trying to solve? Can we simplify the problem statement? Can you break down the problem into its key components? What is the minimal set of operations we need to perform to solve this problem?

Visual Model of the Problem

How to visualize the problem statement for this problem?

Problem Restatement

Could you start by paraphrasing the problem statement in your own words? Try to distill the problem into its essential elements and make sure to clarify the requirements and constraints. This exercise should aid in understanding the problem better and aligning our thought process before jumping into solving it.

Abstract Representation of the Problem

Could you help me formulate an abstract representation of this problem?

Given this problem, how can we describe it in an abstract way that emphasizes the structure and key elements, without the specific real-world details?

Terminology

Are there any specialized terms, jargon, or technical concepts that are crucial to understanding this problem or solution? Could you define them and explain their role within the context of this problem?

Problem Simplification and Explanation

Could you please break down this problem into simpler terms? What are the key concepts involved and how do they interact? Can you also provide a metaphor or analogy to help me understand the problem better?

Constraints

Given the problem statement and the constraints provided, identify specific characteristics or conditions that can be exploited to our advantage in finding an efficient solution. Look for patterns or specific numerical ranges that could be useful in manipulating or interpreting the data.

What are the key insights from analyzing the constraints?

Case Analysis

Could you please provide additional examples or test cases that cover a wider range of the input space, including edge and boundary conditions? In doing so, could you also analyze each example to highlight different aspects of the problem, key constraints and potential pitfalls, as well as the reasoning behind the expected output for each case? This should help in generating key insights about the problem and ensuring the solution is robust and handles all possible scenarios.

Provide names by categorizing these cases

What are the edge cases?

How to visualize these cases?

What are the key insights from analyzing the different cases?

Identification of Applicable Theoretical Concepts

Can you identify any mathematical or algorithmic concepts or properties that can be applied to simplify the problem or make it more manageable? Think about the nature of the operations or manipulations required by the problem statement. Are there existing theories, metrics, or methodologies in mathematics, computer science, or related fields that can be applied to calculate, measure, or perform these operations more effectively or efficiently?

Simple Explanation

Can you explain this problem in simple terms or like you would explain to a non-technical person? Imagine you’re explaining this problem to someone without a background in programming. How would you describe it? If you had to explain this problem to a child or someone who doesn’t know anything about coding, how would you do it? In layman’s terms, how would you explain the concept of this problem? Could you provide a metaphor or everyday example to explain the idea of this problem?

Problem Breakdown and Solution Methodology

Given the problem statement, can you explain in detail how you would approach solving it? Please break down the process into smaller steps, illustrating how each step contributes to the overall solution. If applicable, consider using metaphors, analogies, or visual representations to make your explanation more intuitive. After explaining the process, can you also discuss how specific operations or changes in the problem’s parameters would affect the solution? Lastly, demonstrate the workings of your approach using one or more example cases.

Inference of Problem-Solving Approach from the Problem Statement

Can you identify the key terms or concepts in this problem and explain how they inform your approach to solving it? Please list each keyword and how it guides you towards using a specific strategy or method. How can I recognize these properties by drawing tables or diagrams?

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Could you please provide a stepwise refinement of our approach to solving this problem?
How can we take the high-level solution approach and distill it into more granular, actionable steps?
Could you identify any parts of the problem that can be solved independently?
Are there any repeatable patterns within our solution?

Solution Approach and Analysis

Identify Invariant

What is the invariant in this problem?

Identify Loop Invariant

What is the loop invariant in this problem?

Is invariant and loop invariant the same for this problem?

Identify Recursion Invariant

Is there an invariant during recursion in this problem?

Is invariant and invariant during recursion the same for this problem?

Thought Process

The first thing to observe is that we need to find the minimum speed, which is an optimization problem. The speed range is specified (1 to 107) and we have to find a specific speed which fulfills our conditions. This hints towards the use of a binary search approach for optimization.
Binary search works by choosing a middle value in a sorted list (or a range in this case), then discarding half the list. If the middle value isn’t the target, we can determine if the target is likely to be in the lower or upper half of the list. This reduces our search space and we repeat the process until we find the target.
Here, the ’target’ is the speed that would allow us to reach the office in the given ‘hour’. We need to compute the time it would take to travel all the train distances at a given speed, and see if it fits within the ‘hour’. We start from the middle of the speed range, and adjust our search space based on whether the time taken is less or more than ‘hour’.
It’s important to remember that each train ride can only start at an integer hour mark. This means, if a train ride ends at a non-integer hour, we need to wait until the next integer hour before the next ride can begin. This waiting time needs to be considered in the total time computation.
When computing the time for each train ride at a given speed, we round up the time to the next integer if it’s not already an integer (because the next train can only depart at an integer hour). However, for the last train ride, we don’t need to do this because we can reach the office right after the last train ride ends.

In this code, we first check whether it’s possible to reach the office in the given time, based on the number of train rides. If it’s impossible, we return -1. Then we perform a binary search over the speed range. For each speed (mid), we compute the total time it would take to travel all the train distances. We round up the time for all train rides except the last one. If the total time exceeds ‘hour’, we need a higher speed, so we adjust our search space to the upper half. If not, we adjust our search space to the lower half. We repeat the process until we find the minimum speed that allows us to reach the office in the given time.

This solution makes use of binary search for optimization, and also considers the constraints related to the train departure times. The complexity of this solution is O(n log m), where n is the number of trains and m is the maximum speed. The ’log m’ comes from the binary search, and for each mid-speed value, we traverse all train distances, which gives the ’n’. This is an efficient solution for the problem.

Failed to write working code after several iterations. Working solution from LC discussion:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
import math
class Solution:
    def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
      def can_reach(speed):
        so_far = 0
        for i in range(len(dist) - 1):
            so_far += math.ceil(dist[i] / speed)
            if so_far > hour:
                return False
        return (so_far + (dist[-1] / speed)) <= hour

      l = 1
      r = 10**7
      max_speed = -1
      while l <= r:
          m = (l + r) // 2
          if can_reach(m):
              max_speed = m
              r = m - 1
          else:
              l = m + 1
      return max_speed

This is the solution to a problem that’s related to finding the minimum speed required to reach a destination within a certain time frame. The destinations are represented as an array of distances, and the total time allowed is given in hours.

This problem is solved using binary search and a helper function that verifies if a given speed is sufficient to reach all destinations within the allowed time. Here’s a breakdown of the code:

can_reach(speed): This is a helper function that checks whether it is possible to reach all destinations within the allotted time given a certain speed. It uses a ‘for’ loop to iterate through all the distances (except the last one) and adds the time needed to cover each distance to a running total (so_far). If at any point the total time exceeds the allowed time, the function returns False. For the last distance, it directly adds the time required (distance/speed) to the running total and checks whether the total time is less than or equal to the allowed time.
minSpeedOnTime(dist, hour): This is the main function that employs a binary search to find the minimum speed needed to reach all destinations in time. The search space is defined between 1 and 10^7. If the mid-point speed allows reaching all destinations in time, it is considered as a potential answer (max_speed), and the search continues in the left half (reducing the speed). If the mid-point speed is not sufficient, the search continues in the right half (increasing the speed).
This continues until the search space is exhausted, and the minimum possible speed that allows reaching all destinations in time is returned. The variable max_speed is initially set to -1 to handle cases where no speed allows reaching all destinations in time.

While the code is written in a simple and straightforward manner, the underlying concepts of binary search and time calculation might be tricky for beginners. It’s important to understand these concepts before diving into the code.

Establishing Preconditions and Postconditions

Parameters:
- What are the inputs to the method?
- What types are these parameters?
- What do these parameters represent in the context of the problem?
Preconditions:
- Before this method is called, what must be true about the state of the program or the values of the parameters?
- Are there any constraints on the input parameters?
- Is there a specific state that the program or some part of it must be in?
Method Functionality:
- What is this method expected to do?
- How does it interact with the inputs and the current state of the program?
Postconditions:
- After the method has been called and has returned, what is now true about the state of the program or the values of the parameters?
- What does the return value represent or indicate?
- What side effects, if any, does the method have?
Error Handling:
- How does the method respond if the preconditions are not met?
- Does it throw an exception, return a special value, or do something else?

Problem Decomposition

Problem Understanding:
- Can you explain the problem in your own words? What are the key components and requirements?
Initial Breakdown:
- Start by identifying the major parts or stages of the problem. How can you break the problem into several broad subproblems?
Subproblem Refinement:
- For each subproblem identified, ask yourself if it can be further broken down. What are the smaller tasks that need to be done to solve each subproblem?
Task Identification:
- Within these smaller tasks, are there any that are repeated or very similar? Could these be generalized into a single, reusable task?
Task Abstraction:
- For each task you’ve identified, is it abstracted enough to be clear and reusable, but still makes sense in the context of the problem?
Method Naming:
- Can you give each task a simple, descriptive name that makes its purpose clear?
Subproblem Interactions:
- How do these subproblems or tasks interact with each other? In what order do they need to be performed? Are there any dependencies?

From Brute Force to Optimal Solution

Certainly, let’s break down this problem from the start.

Brute Force Approach

A naive or brute force solution to this problem would be to try all possible speeds for the trains from 1 to the maximum distance in the dist array multiplied by 10^7 (since the problem states that the speed does not exceed 10^7), and then check if you can reach the office in time with the current speed.

For each speed, you would calculate the total time for all train rides (distance/speed), and for all but the last train ride, you would round up the time to the nearest integer to account for waiting times. If the total time is less than or equal to the hour, then you have found a speed that allows you to reach the office in time.

Here’s a Python code snippet that illustrates this brute force approach:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
import math
from typing import List

class Solution:
    def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
        max_speed = max(dist) * 10**7
        for speed in range(1, max_speed + 1):
            total_time = sum([math.ceil(d/speed) for d in dist[:-1]]) + dist[-1] / speed
            if total_time <= hour:
                return speed
        return -1

Inefficiencies of the Brute Force Approach

This brute force solution is simple and straightforward, but it is highly inefficient. Its time complexity is O(nm), where n is the length of the dist array and m is the maximum possible speed. In the worst case scenario, when the maximum distance in dist is 10^5, the time complexity becomes O(10^12), which is prohibitively high for most computing systems. This means that the brute force solution will not work within the time limits for large inputs.

Optimizing the Solution

To optimize the solution, we can use a binary search algorithm to find the minimum speed. Binary search reduces the number of speeds we need to check from m to log(m), significantly improving the time complexity.

In the binary search, the lower bound is 1 and the upper bound is the maximum possible speed. The mid-point of the lower and upper bounds is chosen as the current speed. The total time for all train rides with this speed is calculated. If it’s less than or equal to the hour, the upper bound is updated to the mid-point; otherwise, the lower bound is updated to mid-point + 1.

Here’s the Python code that implements the binary search approach:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
import math
class Solution:
    def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
      def can_reach(speed):
        so_far = 0
        for i in range(len(dist) - 1):
            so_far += math.ceil(dist[i] / speed)
            if so_far > hour:
                return False
        return (so_far + (dist[-1] / speed)) <= hour

      l = 1
      r = 10**7
      max_speed = -1
      while l <= r:
          m = (l + r) // 2
          if can_reach(m):
              max_speed = m
              r = m - 1
          else:
              l = m + 1
      return max_speed

Impact on Time and Space Complexity

The binary search approach has a time complexity of O(n log m), where n is the length of the dist array and m is the maximum possible speed. This is significantly better than the brute force solution, making it feasible for large inputs. The space complexity remains O(1) as no additional data structures are used. The optimization mainly reduces the number of iterations, greatly improving the time efficiency of the solution.

Code Explanation and Design Decisions

Initial Parameters: The initial parameters are dist, which is a list of distances for each train ride, and hour, which is the time we have to get to the office. These inputs represent the problem constraints and they define the amount of time and distances we need to cover.
Primary Loop: The primary loop in this function is the while loop (while l <= r:). This loop is implementing the binary search. Here, the binary search is being used to find the minimum speed that allows you to reach the office on time.
Conditions or Branches: The condition in the loop is if can_reach(m):, which checks if it’s possible to reach the office on time with the current speed (m). If it’s possible, it updates max_speed and tries to find a smaller valid speed in the next iteration. If it’s not possible, it increases the minimum speed (l) for the next iteration.
Parameter Updates: Inside the while loop, we have three parameter updates: max_speed = m, r = m - 1, and l = m + 1. These changes represent the update steps in the binary search algorithm. Depending on whether the current middle speed (m) is valid, we adjust our search range accordingly.
Invariants: The invariant here is the binary search condition l <= r. This condition ensures we continue the search as long as our range is valid. This invariant guarantees that we do not miss any potential answer within the range and helps us to find the minimum valid speed.
Output: The final output is max_speed, which is the minimum speed required to reach the office within the given time. If it’s impossible to reach the office on time even at the maximum speed, the function will return -1, as per the problem’s requirements.

Coding Constructs

High-Level Strategies: The code uses the binary search strategy, which is a divide-and-conquer algorithm. This strategy is used to efficiently search for a solution in a sorted or ordered problem space. Here, it is applied to find the minimum train speed that allows you to reach the office on time.
Purpose of the Code (for a non-programmer): This code finds out the minimum speed that all the trains must travel at in order to get you to your office on time. It considers the time you have, the distances you need to travel, and the rule that each train can only depart at an integer hour mark. It uses a smart strategy to search through possible speeds, starting from the middle and then checking higher or lower speeds depending on whether the current speed is too slow or fast.
Logical Elements: The logical elements used in this code include loops, conditional statements, and functions. A helper function can_reach() is defined within the main function to check if a given speed would allow reaching the office on time. Loops are used for iteration over train rides and for implementing the binary search. Conditional statements are used to check if the current speed is valid and to decide the next step in the binary search.
Algorithmic Approach in Plain English: The code first calculates the possible range of train speeds. It starts by checking the speed in the middle of this range. If this speed allows reaching the office on time, it then checks for any lower speed in the same way. If it doesn’t, it checks for any higher speed. It keeps doing this until it finds the minimum speed that allows reaching the office on time.
Key Steps or Operations: The code primarily performs three operations:
- It calculates the time taken for each train ride at a given speed.
- It adds these times up to check if they fit within the given time limit.
- It uses binary search to efficiently find the minimum valid speed.
Algorithmic Patterns: The key algorithmic pattern used here is binary search. This strategy involves dividing the problem into smaller parts and recursively solving those parts. This code applies binary search to the range of possible speeds to efficiently find the minimum valid speed. It also uses the ‘check and adjust’ pattern where it checks if a solution is valid and adjusts its approach based on the check’s result.

Language Agnostic Coding Drills

Dissection of the code:
- Variables and Arithmetic Operations: The very basic concept is to understand what variables are and how to perform simple arithmetic operations. This is used to calculate the speed, time, etc.
- Arrays/Lists: Here, we have an array of distances that the trains need to travel. Understanding how to store and retrieve data from arrays is essential.
- Loops: The code contains loops to iterate through each train ride and the possible range of speeds. Understanding the concept of loops and how they are controlled is key to implementing this solution.
- Conditional Statements: Conditional statements, such as if/else, are used to control the flow of the code based on certain conditions. In this code, they’re used to check if a given speed is valid and to decide the next step in the binary search.
- Functions: The code defines a helper function can_reach(). Understanding how to define and use functions is necessary to structure the code effectively.
- Binary Search Algorithm: This is an advanced concept that involves understanding how to divide the problem space in half at each step to find a solution efficiently.
List of Concepts in Increasing Difficulty:
- Variables and Arithmetic Operations: This is the most basic concept that any beginner programmer needs to understand.
- Arrays/Lists: Slightly more complex than basic variables, understanding arrays requires understanding of memory allocation and indexing.
- Conditional Statements: This is slightly more advanced as it requires logical reasoning skills to determine the conditions under which code blocks should be executed.
- Loops: Looping is a bit more complex, as it involves understanding of control flow and conditions for termination.
- Functions: Writing functions involves understanding of scopes, arguments, return values, and overall code organization.
- Binary Search Algorithm: This is the most advanced concept in the list. It requires understanding of algorithms, recursion or iteration, and how to apply it to specific problems.
Problem-Solving Approach:
- Understanding the Problem: Start by understanding the problem, the constraints, and the expected output.
- Identifying Variables: Identify what variables will be required. This will include distances, hour, speed etc.
- Implementing the Basic Operations: Perform basic operations like calculating time taken for each train ride at a given speed.
- Using Arrays/Lists: Store the distances in an array and learn to iterate over them.
- Applying Conditional Statements: Apply conditions to check if the calculated time exceeds the given time limit.
- Applying Loops: Use a loop to iterate through each speed from the minimum to the maximum speed.
- Defining and Using Functions: Define a helper function to calculate if a given speed allows you to reach on time.
- Applying Binary Search: Apply the binary search algorithm to find the minimum valid speed more efficiently. This involves choosing the middle speed, then choosing a lower or higher speed depending on whether the current speed is valid or not, and repeating this until the minimum valid speed is found.

Targeted Drills in Python

Alright, let’s break down the code into coding drills.

Variables and Arithmetic Operations:

1
2
3
4
5
6
# Variables and arithmetic operations
a = 5
b = 2
sum_ab = a + b
product_ab = a * b
print(sum_ab, product_ab)  # Output: 7 10

Arrays/Lists:

1
2
3
4
# Working with lists
distances = [1, 3, 2]
print(distances[0])  # Output: 1
print(distances[2])  # Output: 2

Conditional Statements:

1
2
3
4
5
6
7
# Conditional statements
a = 5
b = 2
if a > b:
    print("a is greater than b")  # Output: a is greater than b
else:
    print("b is greater or equal to a")

Loops:

1
2
3
# Loops
for i in range(5):
    print(i)  # Output: 0 1 2 3 4

Functions:

1
2
3
4
5
# Functions
def square(number):
    return number ** 2

print(square(5))  # Output: 25

Binary Search Algorithm:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
# Binary Search
def binary_search(array, target):
    left, right = 0, len(array) - 1
    while left <= right:
        mid = (left + right) // 2
        if array[mid] == target:
            return mid
        elif array[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

numbers = [1, 2, 3, 4, 5]
print(binary_search(numbers, 3))  # Output: 2

With these drills in place, we can start putting them together to solve our initial problem.

Start by defining your input variables (speed, distances, hour).
Use loops and array manipulation to calculate the time it takes to travel each distance at a given speed. This is a good point to integrate the can_reach() function.
Incorporate the binary search algorithm to iteratively adjust the speed until finding the minimum speed that will get you to your destination on time. Make sure to use your conditional statements to correctly adjust the left and right boundaries of your search.
Once you’ve found the correct speed, return that as the output of your solution.

Q&A

how about change “min = 1” to “min = the sum of dist / hour”.

In the given problem, we are essentially finding the minimum speed needed to cover all distances in the given time.

We initiate the speed as 1 because we assume that speed can’t be lower than 1 unit/hour. If we started with “min = sum of dist / hour”, we would be initializing with an average speed required to cover all distances in the given time. This is not necessarily the minimum speed, because some distances might be covered faster than others.

However, we could certainly use “min = sum of dist / hour” as a starting point, especially if we have some knowledge about the data that informs us this would be a sensible starting point. This would not affect the correctness of the solution, but could potentially impact the efficiency. If “min = sum of dist / hour” is closer to the actual minimum speed, the binary search could potentially reach the solution faster.

Remember though, the initial values of “min” and “max” in a binary search are usually selected based on the range of possible solutions. In this case, the speed ranges from 1 unit/hour (the minimum possible speed) to 10^7 units/hour (a high value to ensure the maximum speed is within this range). This ensures that the binary search checks all possible speeds.

You don’t have to search all of 1 through 10^7. Instead, first check if hour <= len(dist)-1. If not you can immediately return -1. Moreover, you can set your initial bounds to be sum(dist)//hour and int(max(max(dist), dist[-1]/(hour-(n-1)))

You’re correct. Implementing these checks could optimize our solution.

If the given hour is less than len(dist) - 1, we can’t reach our destination, even if we traveled at maximum speed. This is because we’re assuming that we can’t travel to the next city instantly and each travel between cities will take at least 1 hour. Thus, we can return -1 immediately.
To establish the initial bounds for the binary search:
a. The lower bound could be sum(dist)//hour. This represents the average speed required to cover all distances in the given time. This is a sensible starting point, as the actual speed can’t be lower than this.
b. The upper bound could be max(max(dist), dist[-1]/(hour-(n-1))). This represents the maximum speed required to reach the last city in time after reaching all other cities in the minimum possible time. We can’t have a speed lower than this, as we won’t reach the last city in time.

These changes won’t affect the correctness of the solution, but could make the binary search more efficient, as we would be searching within a narrower and more targeted range.

You can get a small speedup by avoiding doing as much floating point division. You can check the integer truncated division and add one if the numbers aren’t evenly divisible to get the ceiling. You only need to do floating point division on the final element in the distance array. My time required function as example:

Sure, you’re right. When it comes to optimization, even a little bit counts. We can reduce the number of floating point divisions by checking if the numbers aren’t evenly divisible and add one in such cases. Here’s how you can modify the can_reach function to use integer division:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
def can_reach(speed):
    total_time = 0
    for i in range(len(dist) - 1):
        time = dist[i] // speed
        if dist[i] % speed != 0:  # if the numbers aren't evenly divisible
            time += 1
        total_time += time
        if total_time >= hour:
            return False
    total_time += dist[-1] / speed  # floating point division only for the final element
    return total_time <= hour

With this change, most of the divisions in the can_reach function are now integer divisions, which could give a small speedup. This alteration, however, doesn’t change the overall time complexity of the algorithm.

The provided code defines a solution to a problem that’s related to finding the minimum speed required to reach a destination within a certain time frame. The destinations are represented as an array of distances, and the total time allowed is given in hours.

This problem is solved using binary search and a helper function that verifies if a given speed is sufficient to reach all destinations within the allowed time. Here’s a breakdown of the code:

can_reach(speed): This is a helper function that checks whether it is possible to reach all destinations within the allotted time given a certain speed. It uses a ‘for’ loop to iterate through all the distances (except the last one) and adds the time needed to cover each distance to a running total (so_far). If at any point the total time exceeds the allowed time, the function returns False. For the last distance, it directly adds the time required (distance/speed) to the running total and checks whether the total time is less than or equal to the allowed time.
minSpeedOnTime(dist, hour): This is the main function that employs a binary search to find the minimum speed needed to reach all destinations in time. The search space is defined between 1 and 10^7. If the mid-point speed allows reaching all destinations in time, it is considered as a potential answer (max_speed), and the search continues in the left half (reducing the speed). If the mid-point speed is not sufficient, the search continues in the right half (increasing the speed).
This continues until the search space is exhausted, and the minimum possible speed that allows reaching all destinations in time is returned. The variable max_speed is initially set to -1 to handle cases where no speed allows reaching all destinations in time.

Remember, while the code is written in a simple and straightforward manner, the underlying concepts of binary search and time calculation might be tricky for beginners. Therefore, it’s important to understand these concepts before diving into the code.

10^7 is a pretty ridiculous upper limit most of the time. I got top tier times with a few optimizations - corner cases of 1 train per hour or less, or minimal speed needed, then used the results from calculating the max speed that mattered as the upper limit for binary search instead.

Indeed, starting the binary search from a more reasonable upper limit can improve the efficiency of the solution. The maximum speed required is usually much lower than 10^7.

The maximum speed is determined by the largest distance to be covered or the overall distance that needs to be covered within the given hours, whichever is greater. As you noted, it’s a good idea to handle corner cases where the total hours is less than or equal to the number of trains. Here is how the revised code might look:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
import math
class Solution:
    def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
      def can_reach(speed):
        total_time = 0
        for i in range(len(dist) - 1):
            time = dist[i] // speed
            if dist[i] % speed != 0: 
                time += 1
            total_time += time
            if total_time >= hour:
                return False
        total_time += dist[-1] / speed
        return total_time <= hour
      
      if hour <= len(dist) - 1:  # corner case of 1 train per hour or less
          return -1
      l = 1
      r = max(max(dist), int(sum(dist) / (hour - len(dist) + 1)))  # optimized upper limit
      max_speed = -1
      while l <= r:
          m = (l + r) // 2
          if can_reach(m):
              max_speed = m
              r = m - 1
          else:
              l = m + 1
      return max_speed

This optimized version makes the binary search more efficient by reducing the search space from the outset.

If we use (dist[i] + speed- 1) / speed instead of Math.Ceil(t) , the float is slower than integer, so we can faster.

You’re correct. Using integer division can be faster than using floating-point division. This is because integer division is a simpler operation for the computer to perform.

The expression (dist[i] + speed - 1) / speed effectively rounds up the division result to the nearest integer, similar to math.ceil(dist[i] / speed). The addition of speed - 1 before the division ensures that any fraction part of the division will result in an additional increment of the quotient, thus mimicking the ceiling function.

Here’s how the can_reach function would look with this change:

1
2
3
4
5
6
7
8
9
def can_reach(speed):
    total_time = 0
    for i in range(len(dist) - 1):
        time = (dist[i] + speed - 1) // speed
        total_time += time
        if total_time >= hour:
            return False
    total_time += dist[-1] / speed
    return total_time <= hour

This adjustment could slightly improve the speed of the function.

Possible optimization to initial search space: If hour <= n - 1, then it is not possible to be on time. No matter how fast our speed is, the first n-1 trains will take at least one hour each, so n-1 hours in total. Note the constraints state that there will be at most two digits after the decimal point in hour. So suppose hour = (n-1) + 0.01. It’s possible to make it on time, but we would need our speed to be at least 100 * dist[n-1]. In the worst case, the last train would also be the largest distance, i.e. dist[n-1] == max(dist). So the right end of our initial search space could be set to max(dist) * 100 instead of 10^7. The constraints state dist[i] <= 10^5, so at most right is still initially 10^7, but it could be smaller.

You’re correct. The speed required to reach the final station on time in the absolute minimum possible time would be dist[n-1] * 100. The reason for this is, as you pointed out, the first n-1 train journeys will each take a minimum of one hour regardless of the speed. Thus, if we have hour = (n-1) + 0.01 hours, we have exactly 0.01 hours left to cover the last journey. As 0.01 hours is 1/100th of an hour, we would need our speed to be 100 times the distance of the last journey, dist[n-1], to cover it in time.

Adjusting the upper bound of the search space for binary search to max(dist) * 100 could help to reduce the search space and speed up the computation. Here’s how the updated binary search initialization would look:

1
2
3
# binary search initialization
l = 1
r = max(dist) * 100

Keep in mind that this adjustment will only make a significant difference in performance when max(dist) is significantly less than 10^5, which is the upper limit for dist[i] according to the problem’s constraints.

Now we have to find a speed x such that total time taken is less than the time given in problem, if I take too less speed then, it will take much more time and if we take more speed it will take less time but we need to keep speed to minimum

So, lets find timeTaken for speed m, if it taken less time then, it is a candidate for our answer, but less is also possible so here decrease right bound of speed else if time taken is more, then speed was less so we need more speed so increment the lower bound

That’s correct. The problem boils down to finding the minimum speed that allows us to reach all stations within the specified hour limit. The technique applied here is binary search.

The algorithm applies binary search on the speed, which starts from a lower limit (l) of 1 and an upper limit (r) of max(dist) * 100. For each iteration, we calculate a mid-value (m) which is the average of l and r.

Then, we calculate timeTaken for speed m. This is done by iterating over the dist array and adding the time taken for each journey to timeTaken. For the last journey, we do not need to take the ceiling as it does not need to wait for an hour before departure.

If timeTaken for speed m is less than or equal to the hour, we set our upper limit to m and continue the binary search with the updated bounds. If timeTaken for speed m is more than the hour, we increase the lower limit to m + 1.

The binary search continues until the lower limit is greater than the upper limit. The upper limit will be our answer, as we are looking for the minimum speed that allows us to reach all stations within the specified hour limit.

This binary search optimization enables us to find the minimum speed in logarithmic time complexity, which is an improvement over a linear scan of all possible speeds.

same idea, but forget there is a max on the statement. how about change “min = 1” to “min = the sum of dist / hour”.

Indeed, your suggestion of setting the lower bound of the binary search to sum(dist) / hour seems logical because that’s the average speed required to cover the total distance in the given time.

However, it’s important to remember that this average speed might not be feasible due to the constraint that you can’t split the journey for a single distance element in the dist list. Each element in dist must be travelled at a constant speed.

Therefore, even if the average speed seems low, you might actually need a higher minimum speed to ensure that each individual journey in dist can be completed within the time limit. So, in this case, starting the lower bound from 1 can cover this scenario and ensure we find the true minimum speed.

It’s crucial to keep the constraints and requirements of the problem in mind when deciding on initial parameters for problem-solving strategies like binary search.

In Java, 1.03 + 1.0 = 2.0300000000000002. So my code failed at the last test: dist: [4,2,3] hour: 2.03 The answer is 100. My code yield 101. All because the 1.03 + 1.0 = 2.0300000000000002 in Java.

The issue you’re facing stems from floating-point arithmetic precision problems, which are prevalent in most programming languages, not just Java.

Floating-point numbers are represented as binary fractions in computer memory. However, many decimal fractions cannot be represented exactly as binary fractions, which can lead to small rounding errors.

In your case, the sum of 1.03 and 1.0 results in 2.0300000000000002 due to such precision issues.

To work around this, you can use a small threshold to account for the imprecision. Instead of checking whether the total time equals hour, you can check if the absolute difference between the total time and hour is smaller than a small threshold, say 1e-8.

This way, if the total time is 2.0300000000000002 and hour is 2.03, the difference would be 2e-16, which is less than the threshold, and the two numbers will be considered as equal. This is a common practice when dealing with floating-point numbers in programming.

Here’s an illustration of how to adjust the code (in Python) for reference:

1
2
3
4
5
6
7
8
9
def can_reach(speed):
    total_time = 0
    for i in range(len(dist) - 1):
        total_time += math.ceil(dist[i] / speed)
    total_time += dist[-1] / speed

    # return total_time == hour 
    # replace with
    return abs(total_time - hour) < 1e-8

In Java, you could use Math.abs(total_time - hour) < 1e-8.

This adjustment should help avoid the problem caused by floating-point precision issues.

how to determine that it’s a question of binary search?

The problem “Minimum Speed to Arrive on Time” falls into a category of problems often called “search problems”, where the goal is to find a specific value that satisfies certain conditions. Binary search is a powerful technique to solve such problems, especially when the following conditions are met:

Monotonicity: If the problem has monotonic properties, it’s a strong indication that binary search can be a viable strategy. Monotonicity here means that if a value x satisfies the condition, then either all values above x (in case of monotonic increasing) or all values below x (in case of monotonic decreasing) also satisfy the condition. In this problem, if a certain speed is sufficient to deliver the packages on time, then any speed faster than that will also suffice.
Search Space is Ordered: Binary search operates on a sorted or ordered sequence. If the search space of the problem is or can be arranged in an ordered sequence, binary search becomes a potential approach. Here, the possible speeds form a clear range (from 1 to 10^7).
Existence of a Check Function: If we can quickly check whether a candidate solution is valid or not, we can use binary search. In this problem, we quickly check whether a candidate speed is sufficient or not.
Cannot Find an Optimal Solution Through Other Means: Sometimes, the nature of the problem does not lend itself to other algorithmic paradigms like dynamic programming or greedy algorithms. In such cases, if the problem matches the first three points, binary search is a reasonable option.

While these conditions can hint towards binary search as a potential solution, the final decision always depends on the specifics of the problem and how well one understands the underlying concepts. Recognizing binary search problems comes with practice and experience in problem solving.

The range of possible values here is 1 - 10^7. In binary search, we have to find answer from a sorted array. Here we are left of finding answer from a sorted range. That’s why it can be done using binary search.

Yes, your understanding is correct. In this problem, the range of possible speeds (1 to 10^7) is our “sorted array”. This sorted range has a property known as “monotonicity”, which is key to applying binary search.

Monotonicity means that if a certain speed is sufficient to deliver all packages within the given hour, any speed higher than that will also be sufficient. Hence, if we find a speed that works, we can eliminate all higher speeds from our search. Similarly, if a speed doesn’t work, we can eliminate it and all lower speeds from our search.

Binary search exploits this property by always choosing the middle speed in our current range, and using the result to cut our search space in half. This process repeats until we find the smallest possible speed that will work, thereby giving us the most efficient solution.

Usually in binary search we make our max = mid -1 when our current > target , but why we are doing opposite here. I am confused.

int binarySearch(int arr[], int l, int r, int x)
{
    if (r >= l) {
        int mid = l + (r - l) / 2;
        if (arr[mid] == x)
            return mid;
        if (arr[mid] > x)
            return binarySearch(arr, l, mid - 1, x);
        return binarySearch(arr, mid + 1, r, x);
    }

The standard binary search function you’ve given is used to find a specific value within a sorted array. Here, if the value at the middle index is greater than the target value, we adjust the right boundary to mid - 1, effectively searching in the left half. If the middle value is less than the target, we adjust the left boundary to mid + 1, to search in the right half.

In our problem, however, we’re not searching for a specific value, but the lowest possible speed that still meets the condition of delivering all packages within the given time. Here, the binary search is used to find the minimum value that meets a certain condition within a sorted range.

If the speed at ‘mid’ allows all deliveries within time, we adjust the right boundary to ‘mid’, because it’s possible there may be a lower speed that also fulfils the condition, and we want the minimum such speed. If the speed at ‘mid’ does not allow for all deliveries, we adjust the left boundary to ‘mid + 1’, because we need a higher speed.

Thus, while the binary search structure remains the same, its implementation can vary depending on the specific problem requirements.

for n-1 index, sum = sum + dist[n-1]/i. Because the time taken by last train need not be converted into Integer. please any one explain (why ?)

The condition of the problem states that only non-integer hours will be counted as full hours for all trains except for the last one.

Let’s say we have 3 trains with distances 5, 10, and 15, and our speed is 10 units/hour.

For the first two trains, even though each takes exactly 0.5 and 1 hour to reach their destinations at this speed, the time is rounded up to the nearest whole number, i.e., to 1 and 2 hours respectively.

However, for the last train, the travel time is exactly 1.5 hours. This time we do not round up. The last train is the only exception in this problem where we don’t need to round up the time taken, so we can leave it as a float.

So the total time taken in this example would be 1 (first train) + 2 (second train) + 1.5 (third train) = 4.5 hours.

The non-rounded time for the last train provides a more accurate measure of the total time taken, and it’s this precision that can help ensure we don’t overestimate the minimum speed needed.

Isn’t the max value you are using here 10^8? max = 100000000…. It should be 10^7…

You’re right, the maximum speed should indeed be set to 10^7 as per the problem constraints. It’s important to correctly set the limits for binary search based on the problem’s constraints. In this case, as the speed of the train can range from 1 to 10^7, these should be the limits used for the binary search. Mistakenly setting it to 10^8 might cause unnecessary iterations and could potentially lead to incorrect results or time limit exceeded errors. So, the correct code should have the maximum speed as 10^7. Thank you for pointing out this mistake!

Why we can’t initialize sum = (double) dist[n-1] / mid; instead sum = 0; ? It fails for “[4,2,3]” case. Also, when to use mid = (min + max) / 2 and when to use mid = min + (max - min) / 2 in binary search? I always fails at this point. Which is correct way to find mid?

Regarding the initialization of sum, starting with sum = (double) dist[n-1] / mid; would be incorrect because the sum variable is intended to accumulate the total time taken by all the trains except the last one. This calculation takes place within the loop. If we were to initialize sum as you suggested, we’d be starting off with the time taken by the last train, and then adding the time taken by all the other trains in the loop. This would mess up our calculations because we would be considering the last train twice.
As for the calculation of the mid value in binary search, both mid = (min + max) / 2 and mid = min + (max - min) / 2 are mathematically equivalent. However, the second version is often used to prevent potential integer overflow.
In some programming languages, if min and max are both large, min + max could overflow the maximum value that can be held by the variable type, leading to incorrect results. By using mid = min + (max - min) / 2, we avoid this potential issue because (max - min) / 2 will always be a value that can be safely added to min without risk of overflow.
In Python, you generally don’t have to worry about integer overflow, so you can use either version. But if you’re coding in a language like C, C++, or Java, you should generally use mid = min + (max - min) / 2 to be safe.

Important points to look for in the question ——————————————– + We must take n trains in sequential order. + dist[i] is the maximum-distance the ith train can cover. + Each train can only depart at an Integer hour. => If you choose ith train that can cover maximum dist[i] with a speed ‘v’ then timen by it is can be calculated by

        time = dist[i] / v (can be floting point number as well)
        
        [As we know that time = distance / speed]
        
        To take the next train, we must have to wait for next nearest integer
        hour from "time"[calculated for current train].

/* Approach:1 By linearly picking up the speed ‘v’ in the range [1 <= v <= 10^7] ——————————————————————————— Try every single speed ‘v’ in the range [1 <= v <= 10^7] and check with this speed if all the train travel will commute you to the offic on time or NOT. Total time taken must be <= given-hour

        Formula used to calculate the time would be : distance / speed
        
    TC : O(len(dist) * log(max_allowed_speed)), where max_allowed_speed = 10^7
    SC : O(1)
    
    The moment we found out the first such speed(v) would be the our answer.
    
Approach:2 By applying Binary Search on the speed 'v' in the range [1 <= v <= 10^7]
====================================================================================

    Instead picking up the speed 'v' linearly in the increament of 1, we can choose
    it on the fly while applying binary search.
    
    Rest of the operation would be the same as approach 1.
    
    TC : O(len(dist) * log(max_allowed_speed))
    SC : O(1)
    

NOTE: Least total_time_that_we_can_obtaine_is : len(dist) - 1. [Obvious thing. Think why??]
        Therefore, we can say that if given_hour < (len(dist) - 1) then answer will never exists.

Let’s assume our Predicate function is “isPossibleToReachOffice(dist, time_to_reach, speed)” Returns “True[T]”. If it’s pssible to reach to the office within given “hour”, if all the train travel at minimum speed “speed(v)”. Returns “False[F]”. Otherwise. (Means NOT possible to reach to the office with this min speed)

Possible Output Configurations of our above Predicate function would be
------------------------------------------------------------------------

Type:1 T T T T F F F .....  [At least 1 "T" followed by many "F"]

Type:2 T T T T ...........  [All "T"]

Type:3 F F F F F .........  [All "F"]   (Answer NOT exists, Needs to return some value to indicate it)

Goal: To find the speed(v) corresponding to the first occurrence of "True[T]", if exists.


NOTE: Speed corresponding to the first occurrence of "True[T]" would be the optimal minimum
        speed at which all the train must have to travel then only we can reach to the office
        within given "hour". (total_time_needed_to_take_n_trains <= hour)

In this problem, I came across a weird error. While calculating the time taken to reach office (in this code referred by variable “sum”) for a given train speed, I used: Math.ceil( (double) (dist[i] / speed) ) and it gave error for second example case and works fine if I do Math.ceil( ( (double) dist[i] ) / speed ) ).

Basically, there seems to be a difference between 1.000 / 3 and 1 / 3. I can not generalize the error and hence could not figure out the root cause. Please help.

I faced a lot of off-by-one errors. How did you determine the conditions: min <= max (and not min < max), min = mid+1; (and not min = mid;) and max = mid-1; (and not max = mid;)? What was the invariant that you used? Greatly appreciate your help!

What will be the time complexity for this solution?